Measuring Central Tendency

read mean document

\begin{equation*} \text{Sum of all elements} = X_1 + X_2 + X_3 + X_4 + X_5 + . . . + X_n \\ \end{equation*}

위의 sum of all elements는 아래와 같이 표현된 수 있다.

\begin{equation*} X_1 + X_2 + X_3 + X_4 + X_5 + . . . + X_n = \sum\limits_{i=1}^{n} X_i \end{equation*}

위의 sum of all elements를 n으로 나누는 것이 평균

\begin{equation*} \frac{\sum\limits_{i=1}^{n} X_i}{n} \end{equation*}

이것을 우리는 흔히 “무”라고 부른다.
\begin{equation*} \mu = \frac{\sum\limits_{i=1}^{N} X_i}{N} \end{equation*}

위처럼 frequency로 정리된 표는 아래와 같이 계산할 수 있다.

\begin{equation*} \begin{split} \mu = \frac{\sum\limits_{}^{} \text{fx}}{\sum{\text{f.nothing}}} = \frac{1 \text{x} 19 + 3 \text{x} 20 + 1 \text{x} 21}{5} \end{split} \end{equation*}

see median document

{19, 19, 20, 20, 20, 21, 21, 100, 102}

{19, 20, 20, 20, 21, 21, 100, 102}

see also quartile document

one <- c(1,1,1,1,2,2,2,2,2,2,3,3,3,3,4,4,4,4,5,5,5,6,6,7,8)
two <- c(1,4,6,6,8,8,8,9,9,9,9,10,10,10,10,11,11,11,11,11,12,12,12,12,12)

> median(one)
[1] 3
> mean(one)
[1] 3.44

> median(two)
[1] 10
> mean(two)
[1] 9.28

> hist(one)
> hist(two)
> par(mfrow=c(1,2)) 
> hist(one)
> hist(two)

par(mfrow=c(1,2)) 
hist(one)
abline(v=mean(one), col="blue", lty=1)
abline(v=median(one), col="red", lty=2)
legend(3.8,9.5, legend=c("mean", "median"), col=c("blue", "red"), lty=1:2, cex=.8)
hist(two)
abline(v=mean(two), col="blue", lty=1)
abline(v=median(two), col="red", lty=2)
legend(.5,9.5, legend=c("mean", "median"), col=c("blue", "red"), lty=1:2, cex=.8)
par(mfrow=c(1,1))  

quartile 명령어는 r에서는 없다. 대신 quantile을 사용한다. 둘은 비슷하나 quantile은 4분위 이상의 것을 한다.

R에서

temp <- c(19, 19, 20, 20, 20, 21, 21, 100, 102)

> quantile(temp)
  0%  25%  50%  75% 100% 
  19   20   20   21  102 
> quantile(temp, prob=1/4)
25% 
 20 
> quantile(temp, prob=c(1/4,2/4))
25% 50% 
 20  20 
> quantile(temp, prob=seq(0,1,1/4))
  0%  25%  50%  75% 100% 
  19   20   20   21  102 
> quantile(temp, prob=seq(0,1,1/2))
  0%  50% 100% 
  19   20  102 
> median(temp)
[1] 20
> 
> quantile(temp, prob=seq(0,1,1/8))
   0% 12.5%   25% 37.5%   50% 62.5%   75% 87.5%  100% 
   19    19    20    20    20    21    21   100   102 

1. Get mean, median for each data set.
2. State how each is skewed.

> x <- c(1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 6, 6, 7, 8)
> x
 [1] 1 1 1 1 2 2 2 2 2 2 3 3 3 3 4 4 4 4 5 5 5 6 6 7 8
> 

> median(x)
[1] 3
> mean(x)
[1] 3.44
> min(x)
[1] 1
> max(x)
[1] 8
> quantile(x)
  0%  25%  50%  75% 100% 
   1    2    3    5    8 
> 

{1, 1, 1, 2, 2, 2, 2, 3, 3, 31, 31, 32, 32, 32, 32, 33, 33, 33}

{1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 31, 31, 32, 32, 32, 32, 33, 33, 33}

{1, 1, 1, 2, 2, 2, 2, 2, 3, 31, 31, 31, 32, 32, 32, 32, 33, 33, 33}

a <- c(1, 1, 1, 2, 2, 2, 2, 3, 3, 31, 31, 32, 32, 32, 32, 33, 33, 33)
b <- c(1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 31, 31, 32, 32, 32, 32, 33, 33, 33)
c <- c(1, 1, 1, 2, 2, 2, 2, 2, 3, 31, 31, 31, 32, 32, 32, 32, 33, 33, 33)
a
b
c
length(a)
length(b)
length(c)
mean(a)
mean(b)
mean(c)
median(a)
median(b)
median(c)

# output 
> a <- c(1, 1, 1, 2, 2, 2, 2, 3, 3, 31, 31, 32, 32, 32, 32, 33, 33, 33)
> b <- c(1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 31, 31, 32, 32, 32, 32, 33, 33, 33)
> c <- c(1, 1, 1, 2, 2, 2, 2, 2, 3, 31, 31, 31, 32, 32, 32, 32, 33, 33, 33)
> a
 [1]  1  1  1  2  2  2  2  3  3 31 31 32 32 32 32 33 33 33
> b
 [1]  1  1  1  2  2  2  2  2  3  3 31 31 32 32 32 32 33 33 33
> c
 [1]  1  1  1  2  2  2  2  2  3 31 31 31 32 32 32 32 33 33 33
> length(a)
[1] 18
> length(b)
[1] 19
> length(c)
[1] 19
> mean(a)
[1] 17
> mean(b)
[1] 16.21053
> mean(c)
[1] 17.68421
> median(a)
[1] 17
> median(b)
[1] 3
> median(c)
[1] 31
> 

## answer 
n <- 3+2+2 +3+4+4
n 
mean.duckc <- 17
median.duckc <- 17
tot <- mean.duckc * n  
# 2 :: 4 인것은 이미 알고 있음
known.sum <- (1*3)+(2*4)+(3*2)+(31*2)
tot - known.sum 
# 227 = (32*4) + (33*3) 이므로 
duckc <- c(1,1,1,2,2,2,2,3,3,31,31,32,32,32,32,33,33,33)
mean(duckc)
median(duckc)

## how do they change with 2 added and 31 added
duckc.2.added <- c(1,1,1,2,2,2,2,2,3,3,31,31,32,32,32,32,33,33,33)
mean(duckc.2.added)
median(duckc.2.added)

duckc.31.added <- c(1,1,1,2,2,2,2,3,3,31,31,31,32,32,32,32,33,33,33)
mean(duckc.31.added)
median(duckc.31.added)

see mode document

modality
bimodal
trimodal
multimodal
unimodal

It works for both number measured and category measured data.

Mean
\begin{eqnarray*} \mu & = & \frac {\Sigma {X_{i}}} {N} \\ & = & 50000 \\ \end{eqnarray*}
\begin{eqnarray*} \text {2000 raised to the mean} & = & \frac {\sum{(X_{i} + 2000)}} {N} \\ & = & \frac {\sum{(X_{i})} + \sum{(2000)}} {N} \\ & = & \frac {\sum{(X_{i})}} {N} + \frac {N * (2000)} {N} \\ & = & 50000 + 2000 \\ & = & 52000 \end{eqnarray*}
Median
모든 연봉에 2000불을 더하는 것이므로 median 값의 위치는 변하지 않는다. 따라서
\begin{eqnarray*} \text{Original Median} (20000) + 2000 & = & 22000 \end{eqnarray*}
Mode
모든 연봉에 2000불을 더하는 것이므로 mode 값은 원래 모드값인 10000 불에 2000불을 더한 값이다. 따라서
\begin{eqnarray*} \text{Original Mode} (10000) + 2000 & = & 12000 \end{eqnarray*}

Mean
\begin{eqnarray*} \text {10% raised to the mean} & = & \frac {\sum{(1.1 * X_{i})}} {N} \\ & = & \frac {1.1 * \sum{(X_{i})}} {N} \\ & = & 1.1 * \frac {\sum{(X_{i})}} {N} \\ & = & 1.1 * 50000 \\ & = & 55000 \end{eqnarray*}
Median
모든 연봉에 10%를 더하는 것이므로 median 값의 위치는 변하지 않는다. 따라서
\begin{eqnarray*} \text{Original Median} (20000) * 1.1 & = & 22000 \end{eqnarray*}
Mode
모든 연봉에 10%를 더하는 것이므로 mode 값은 원래 모드값인 10000 불에 1.1을 곱한 값이다. 따라서
\begin{eqnarray*} \text{Original Mode} (10000) * 1.1 & = & 11000 \end{eqnarray*}